Horizontal elastic collision between a ball and a block attached to a spring?

In summary, the problem involves a 0.140 kg block attached to a spring with k = 25 N/m, and a 0.023 kg ball thrown horizontally with a speed of 5.2 m/s towards the block. The question asks for the ball's speed immediately after the perfectly elastic collision. To solve the problem, conservation of momentum and conservation of kinetic energy are used, resulting in two equations and two unknown variables. The length of the compressed spring and use of F = -k(change s) are also mentioned, but not necessary for solving the problem.
  • #1
offbeatjumi
28
0

Homework Statement



The problem states: A 0.140 kg block on a frictionless table is firmly attached to one end of a spring with k = 25 N/m. The other end of the spring is anchored to the wall. A 0.023 kg ball is thrown horizontally toward the block with a speed of 5.2 m/s. What is the ball's speed immediately after the collision (if the collision is perfectly elastic)?

Homework Equations



F(spring) = -k(change in s)
K(f) + U(f) = K(i) + U(i) ...---> K = (1/2)mv^2 ...U = (1/2)k(change s)^2

The Attempt at a Solution



The one thing that prevents me from solving the problem is that I am unable to find the length of the compressed spring, and I also do not know how to use F = -k(change s) in this problem. I know that since this is perfectly ellastic, I should use conservation of momentum or conservation of energy, but I always come out with two unknown variables... the final velocity of the ball and usually, change in s. Any pointers? Thanks so much.
 
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  • #2
I could be wrong on this, but immediately after the collision the spring has barely moved at all. Therefore it is just an elastic collision between a ball and a block. Pretend the spring isn't there.

If there is a later question asking the amplitude of the oscillations, then you will know the initial velocity of the block from the previous elastic equation. And can solve for the amplitude.
 
  • #3
Use conservation of momentum and conservation of kinetic energy. Stop the experiment before the spring starts to compress. You get two variables, but you also have two equations.

It's just algebra from there.
 

1. What is a horizontal elastic collision?

A horizontal elastic collision is a type of collision in which two objects, such as a ball and a block attached to a spring, collide with each other horizontally and bounce off each other due to the elastic properties of the objects.

2. How does a horizontal elastic collision differ from other types of collisions?

A horizontal elastic collision differs from other types of collisions, such as inelastic collisions, because the objects involved bounce off each other with no loss of kinetic energy. This is due to the elastic properties of the objects and the conservation of energy.

3. What factors affect the outcome of a horizontal elastic collision?

The outcome of a horizontal elastic collision is affected by factors such as the mass and velocity of the objects involved, as well as the elastic properties of the objects.

4. How is momentum conserved in a horizontal elastic collision?

In a horizontal elastic collision, the total momentum of the system (the ball and block) before the collision is equal to the total momentum after the collision. This is due to the conservation of momentum, which states that in a closed system, the total momentum remains constant.

5. What is the formula for calculating the velocities of the objects after a horizontal elastic collision?

The formula for calculating the velocities of the objects after a horizontal elastic collision is:
v1f = ((m1-m2)v1i + 2m2v2i)/(m1+m2)
v2f = ((m2-m1)v2i + 2m1v1i)/(m1+m2)
Where v1f and v2f are the final velocities of the objects, m1 and m2 are the masses of the objects, and v1i and v2i are the initial velocities of the objects before the collision.

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